Problem: $h(x) = -4x^{3}+6x^{2}$ $f(n) = 7n^{2}-n+4(h(n))$ $g(t) = 6t^{2}-6t-5-3(h(t))$ $ f(h(1)) = {?} $
Answer: First, let's solve for the value of the inner function, $h(1)$ . Then we'll know what to plug into the outer function. $h(1) = -4(1^{3})+6(1^{2})$ $h(1) = 2$ Now we know that $h(1) = 2$ . Let's solve for $f(h(1))$ , which is $f(2)$ $f(2) = 7(2^{2})-2+4(h(2))$ To solve for the value of $f$ , we need to solve for the value of $h(2)$ $h(2) = -4(2^{3})+6(2^{2})$ $h(2) = -8$ That means $f(2) = 7(2^{2})-2+(4)(-8)$ $f(2) = -6$